A man fires a shot and hears an echo from a cliff after `2 s`. He walks `85 m` towards the cliff and the echo of a second shot is now heard after `1.4 s`. What is the velocity of sound and how far was the man from the cliff when he first heard the echo ?

Explanation

Let the man stationed at `A` at a distance `d` from the cliff fire a shot. If the cliff is denoted by `C`, then `AC = d` as shown in (Fig. 6.24).
The echo of the first shot will be heard if the sound goes from A to C and comes bak to `A`. In the process, it travels a distance `2 d`.
Since a time of `2 s` is taken by the first echo to be heard,
`2 d = v t = v xx 2 = 2 v` (v is the speed of sound)
or `d = v`...(i)
After walking `85 m`, the man comes to `B` where `BC = (d - 85) m`. For the echo of the second shot to be heard, sound has to go from B to C and back to the man at `B`, i.e., it has to travel a distance `2 BC = 2 (d - 85) m = (2 d - 170) m`. Since echo of the second shot is heard after a time of `1.5 s`,
`(2s - 170) = v xx 1.5 = (3 v)/(2)` ...(ii)
From eqns. (i) and (ii),
`2d - 170 = (3 d)/(2) (v = d)`
or `d = 340 m`
Speed of sound, `v = d = 340 m//s`.
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